-2cos(t)/t^2>=-2/t^2 won't help show that diverges. You're bounding it from the wrong side, your bound is 0<=2cos(t)/t^2<=2/t^2, on [0,1] say. Bounding something from above by something divergent tells you nothing.

You can fix this, but why bother with a change of variables at all? You know what cos(2/x) is doing as x->infinity right?

Here's one way of looking at it. Consider the function f(x) = cos(2/x). Now as [itex] x \rightarrow \infty [/itex], f(x) tends to 1. So, when you calculate the area under the graph from 1 to infinity, can you see that the area blows up to infinity?

In order to prove this rigorously, use some version of the simple ordering property of the integral:
If the integral of a function f (strictly greater than your own integrand) diverges, then your integral diverges as well ("bigger integrand means bigger integral").